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now I guess I have to multiplied 2) by 4 because:-JY BGBGBG-YJ BGBGBG-JY GBGBGB-YJ GBGBGB

4!*4!*2 -(3!*3!*7)*4= result

do you think am I right?

First of all, please post answer choices for PS questions. Also do no reword or shorten the stem.

Next, I wouldn't worry about this question at all as it's out of the scope of the GMAT. On the GMAT combination/probability questions are fairly straightforward and there is no need to spent too much time on this kind of problems.

Anyway:There are 4 boys and 4 girls. We should sit them in a row so that boys and girls are seated alternatively and one particular boy (B') and one particular girl (G') are not in adjacent seats.

1. 4 boys and 4 girls can be seated alternately in 4!*4!*2 ways:BGBGBGBG - 4!*4! (4 boys can be arranged in their slots in 4! ways and similarly 4 girls can be arranged in their slots in 4! ways);GBGBGBGB - 4!*4! (the same here, we just start a row with a girl this time).

2. Let's count total # of ways B' and G' are in adjacent seats:

Consider them to be one unit {B'G'}If we start a row with a boy: *BG*BG*BG* --> we can place {B'G'} in one of the 4 empty slots (notice that we are not violating alternately seating restriction in this case) and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 4*3!*3!;If we start a row with a girl: G*BG*BG*B --> we can place {B'G'} in one of the 3 empty slots and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 3*3!*3!;Total for the unit {B'G'}: 4*3!*3!+3*3!*3!=7*3!*3!

The same # of ways if we consider unit to be {G'B'};

Therefore total # of ways where B' and G' are in adjacent seats is 2*7*3!*3!.

3. Finally, {total # of ways where B' and G' are not in adjacent seats and boys and girls are seated alternatively} = {total # of ways where boys and girls are seated alternatively} - {total # of ways where B' and G' are in adjacent seats} = 4!*4!*2 - 2*7*3!*3! = 648.

now I guess I have to multiplied 2) by 4 because:-JY BGBGBG-YJ BGBGBG-JY GBGBGB-YJ GBGBGB

4!*4!*2 -(3!*3!*7)*4= result

do you think am I right?

First of all, please post answer choices for PS questions. Also do no reword or shorten the stem.

Next, I wouldn't worry about this question at all as it's out of the scope of the GMAT. On the GMAT combination/probability questions are fairly straightforward and there is no need to spent too much time on this kind of problems.

Anyway:There are 4 boys and 4 girls. We should sit them in a row so that boys and girls are seated alternatively and one particular boy (B') and one particular girl (G') are not in adjacent seats.

1. 4 boys and 4 girls can be seated alternately in 4!*4!*2 ways:BGBGBGBG - 4!*4! (4 boys can be arranged in their slots in 4! ways and similarly 4 girls can be arranged in their slots in 4! ways);GBGBGBGB - 4!*4! (the same here, we just start a row with a girl this time).

2. Let's count total # of ways B' and G' are in adjacent seats:

Consider them to be one unit {B'G'}If we start a row with a boy: *BG*BG*BG* --> we can place {B'G'} in one of the 4 empty slots (notice that we are not violating alternately seating restriction in this case) and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 4*3!*3!;If we start a row with a girl: G*BG*BG*B --> we can place {B'G'} in one of the 3 empty slots and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 3*3!*3!;Total for the unit {B'G'}: 4*3!*3!+3*3!*3!=7*3!*3!

The same # of ways if we consider unit to be {G'B'};

Therefore total # of ways where B' and G' are in adjacent seats is 2*7*3!*3!.

3. Finally, {total # of ways where B' and G' are not in adjacent seats and boys and girls are seated alternatively} = {total # of ways where boys and girls are seated alternatively} - {total # of ways where B' and G' are in adjacent seats} = 4!*4!*2 - 2*7*3!*3! = 648.

Re: Find the ways in which 4 boys and 4 girls can be seated alternatively
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04 Sep 2018, 04:12

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